Explanation
We are asked to solve the given quadratic equation
[tex]2x^2+3x-20=0[/tex]To do so, we will have to find the factors of the product of the coefficient of the first and last term whose sum is 3
Thus
[tex]2\times-20=-40[/tex]So the factor of -40 which also gives a sum of 3 will be
[tex]8\text{ and -5}[/tex]Thus
[tex]\begin{gathered} 2x^2+8x-5x-20=0 \\ 2x(x+4)-5(x+4)=0 \\ (2x-5)(x+4)=0 \end{gathered}[/tex]We will have the answer will be
[tex](2x-5)(x+4)[/tex]Simplifying further
we will have the zeros to be
[tex]\begin{gathered} 2x-5=0 \\ x=\frac{5}{2}=2.5 \\ \\ \end{gathered}[/tex]Also
[tex]\begin{gathered} x+4=0 \\ x=-4 \end{gathered}[/tex]So the answers are:
x = -4 and x= 5 over 2
